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Radius and interval of convergence, endpoint tests and series sums with step-by-step solutions
Find the interval of convergence of sum (x-2)^n / (n * 3^n) from n=1 to infinity
Find the radius of convergence of sum 2^n x^n / n!
Evaluate sum (x/2)^n from n=0 to infinity at x = 1.5
Does sum 1/n^(3/2) converge?

Radius and Interval of Convergence

A power series centred at cc is

∑n=0∞an(x−c)n\sum_{n=0}^{\infty} a_n (x - c)^n

and the question is always the same: for which xx does it converge? The answer is always an interval centred on cc, of half-width RR, the radius of convergence.

Get RR from the ratio test. Form

L=lim⁡n→∞∣an+1(x−c)n+1an(x−c)n∣L = \lim_{n\to\infty} \left| \frac{a_{n+1}(x-c)^{n+1}}{a_n (x-c)^n} \right|

and demand L<1L < 1. Three outcomes:

  • R=0R = 0 — converges only at x=cx = c (e.g. ∑n! xn\sum n!\,x^n)
  • 0<R<∞0 < R < \infty — converges on (c−R, c+R)(c-R,\ c+R), diverges outside
  • R=∞R = \infty — converges for every real xx (e.g. ∑xn/n!\sum x^n/n!)

The assumption people forget: the ratio test says nothing at ∣x−c∣=R|x - c| = R, because there L=1L = 1 exactly. The two endpoints must be tested separately, and they can behave differently from each other.

Testing the Endpoints

Substituting an endpoint turns the power series into an ordinary numerical series, and two standard results usually settle it.

p-series:

∑n=1∞1npconverges iffp>1\sum_{n=1}^{\infty} \frac{1}{n^p} \quad\text{converges iff}\quad p > 1

So ∑1/n3/2\sum 1/n^{3/2} converges and ∑1/n\sum 1/n — the harmonic series, p=1p = 1 — diverges, marginally but definitely.

Alternating series test: ∑(−1)nbn\sum (-1)^n b_n converges if bnb_n decreases monotonically to zero. This is why ∑(−1)n/n\sum (-1)^n/n converges while ∑1/n\sum 1/n does not; the first is conditionally convergent.

Geometric series, worth recognising instantly:

∑n=0∞arn=a1−rfor ∣r∣<1\sum_{n=0}^{\infty} ar^n = \frac{a}{1-r} \quad \text{for } |r| < 1

When it applies: the geometric sum formula needs ∣r∣<1|r| < 1 strictly; at ∣r∣=1|r| = 1 the terms do not shrink and the series diverges by the nnth term test.

Common Mistakes to Avoid

  • Reporting an open interval without checking the endpoints — half the marks on a typical exam question live there.
  • Assuming both endpoints behave the same way — ∑(x−2)n/(n3n)\sum (x-2)^n/(n3^n) converges at one end and diverges at the other.
  • Confusing pp-series with geometric — ∑1/np\sum 1/n^p has nn in the base, ∑rn\sum r^n has nn in the exponent. They converge under completely different conditions.
  • Thinking an→0a_n \to 0 proves convergence — the harmonic series is the standing counterexample. The nnth term test can only prove divergence.
  • Dropping the absolute values in the ratio test — the limit is of a modulus.
  • Forgetting the centre — the interval is centred at cc, not at 00, so a radius of 33 about c=2c = 2 gives (−1,5)(-1, 5).

Examples

Step 1: Ratio test: ∣(x−2)n+1(n+1)3n+1⋅n 3n(x−2)n∣=∣x−2∣3⋅nn+1\left|\dfrac{(x-2)^{n+1}}{(n+1)3^{n+1}} \cdot \dfrac{n\,3^n}{(x-2)^n}\right| = \dfrac{|x-2|}{3}\cdot\dfrac{n}{n+1}
Step 2: As n→∞n \to \infty this tends to ∣x−2∣/3|x-2|/3; require ∣x−2∣/3<1|x-2|/3 < 1, so R=3R = 3
Step 3: Open interval: −1<x<5-1 < x < 5
Step 4: Endpoint x=5x = 5: ∑3n/(n3n)=∑1/n\sum 3^n/(n3^n) = \sum 1/n, the harmonic series — diverges (p=1p = 1)
Step 5: Endpoint x=−1x = -1: ∑(−3)n/(n3n)=∑(−1)n/n\sum (-3)^n/(n3^n) = \sum (-1)^n/n — converges conditionally by the alternating series test
Answer: R=3R = 3, interval of convergence [−1, 5)[-1,\ 5)

Step 1: Ratio: ∣2n+1xn+1(n+1)!⋅n!2nxn∣=2∣x∣n+1\left|\dfrac{2^{n+1}x^{n+1}}{(n+1)!} \cdot \dfrac{n!}{2^n x^n}\right| = \dfrac{2|x|}{n+1}
Step 2: lim⁡n→∞2∣x∣n+1=0<1\lim_{n\to\infty} \dfrac{2|x|}{n+1} = 0 < 1 for every real xx
Step 3: So the series converges everywhere and R=∞R = \infty
Step 4: Matching ∑un/n!=eu\sum u^n/n! = e^{u} with u=2xu = 2x identifies the sum
Answer: R=∞R = \infty, interval (−∞, ∞)(-\infty,\ \infty), and the sum is e2xe^{2x}

Step 1: This is geometric with ratio r=x/2r = x/2, so it converges for ∣x/2∣<1|x/2| < 1, i.e. ∣x∣<2|x| < 2 and R=2R = 2
Step 2: At x=1.5x = 1.5: r=1.5/2=0.75r = 1.5/2 = 0.75, and ∣0.75∣<1|0.75| < 1, so it converges
Step 3: S=a1−r=11−0.75S = \dfrac{a}{1-r} = \dfrac{1}{1 - 0.75}
Step 4: S=1÷0.25=4S = 1 \div 0.25 = 4
Answer: Converges at x=1.5x = 1.5 with sum S=4S = 4

Frequently Asked Questions

Apply the ratio test to |aₙ₊₁(x−c)ⁿ⁺¹ / aₙ(x−c)ⁿ|, take the limit as n → ∞, and set it less than 1. Solving that inequality gives |x − c| < R, and R is the radius. If the limit is 0 the radius is infinite; if it is infinite for every x ≠ c the radius is 0.

Because at |x − c| = R the ratio-test limit equals 1, and the test is inconclusive there. Substitute each endpoint to get a numerical series and apply a p-series, alternating series or comparison test. The two endpoints often give different answers.

The series Σ 1/nᵖ converges when p > 1 and diverges when p ≤ 1. The boundary case p = 1 is the harmonic series, which diverges even though its terms go to zero — the classic warning against the nth term test being used as a convergence proof.

A series converges absolutely if Σ|aₙ| converges, and conditionally if Σaₙ converges but Σ|aₙ| does not. Σ(−1)ⁿ/n is conditionally convergent. Inside its radius of convergence a power series always converges absolutely; conditional convergence only ever shows up at an endpoint.

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